Contracting a symmetric tensor on another tensor The following text was taken from Sean Caroll's General Relativity (page 27 section 1.7):

If we are contracting over a pair of upper indices that are symmetric
on one tensor, only the symmetric part of the lower indices will
contribute, thus,
$X^{(\mu\nu)}Y_{\mu\nu}=X^{(\mu\nu)}Y_{(\mu\nu)}$
regardless of symmetry properties of $Y_{\mu\nu}$

Where the round bracket indicate symmetrization $X^{(\mu\nu)}=\frac{1}{2}(X^{\mu\nu}+X^{\nu\mu})$
This statement doesn't seem obvious to me, so i tried proving it:
$X^{(\mu\nu)}Y_{\mu\nu}$
$=\frac{1}{2}(X^{\mu\nu}+X^{\nu\mu})Y_{\mu\nu}$
$=\frac{1}{2}(X^{\mu\nu}Y_{\mu\nu}+X^{\nu\mu}Y_{\mu\nu})$
$=\frac{1}{2}(X^{\mu\nu}Y_{\mu\nu}+X^{\mu\nu}Y_{\nu\mu})$ (relabel indices on second term)
$=X^{\mu\nu}Y_{(\mu\nu)}$
But it doesn't match the intended equation. Where did it go wrong?
 A: Since you start with the assumption that $X^{\mu \nu}$ is symmetric $\big(\text{aka } X^{\mu \nu} = X^{(\mu \nu)}\big)$, you can use that property a second time in your proof. Observe that because
$$\frac{1}{2}\bigg(X^{\mu\nu} Y_{\mu\nu} + X^{\mu\nu} Y_{\nu\mu}\bigg) = \frac{1}{2}X^{\mu\nu}\bigg( Y_{\mu\nu} +  Y_{\nu\mu}\bigg)$$
and $$X^{\mu\nu} = X^{(\mu\nu)} = \frac{1}{2}\bigg( X^{\mu\nu}+X^{\nu\mu} \bigg)$$
we can write
$$X^{\mu\nu} Y_{\mu\nu} = \frac{1}{4}\bigg( X^{\mu\nu}+X^{\nu\mu} \bigg)\bigg( Y_{\mu\nu} +  Y_{\nu\mu}\bigg)$$
which is what you're trying to show.
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$\textbf{Edit for OP's benefit:}$
To show more explicitly why this works out, consider the substitution $Z^{\mu \nu} = X^{(\mu \nu)}$ where $Z^{\mu \nu}$ is clearly symmetric and the property
$$Z^{\mu \nu} = \frac{1}{2}\bigg(Z^{\mu \nu} + Z^{\nu \mu}\bigg)$$
falls straight out of the definition of symmetry. Observe that
$$ \begin{align} X^{(\mu \nu)} Y_{\mu \nu} &= Z^{\mu \nu}Y_{\mu \nu}\\
&= \frac{1}{2}\bigg(Z^{\mu \nu}Y_{\mu \nu} + Z^{\nu \mu}Y_{\mu \nu}\bigg)\\
&= \frac{1}{2}\bigg(Z^{\mu \nu}Y_{\mu \nu} + Z^{\mu \nu}Y_{\nu \mu}\bigg)\\
&= \frac{1}{2}Z^{\mu \nu}\bigg(Y_{\mu \nu} + Y_{\nu \mu}\bigg)\\
&= Z^{\mu \nu} Y_{(\mu \nu)} \text{ .} \end{align}$$
Back substituting $Z^{\mu \nu} = X^{(\mu \nu)}$, we complete showing the equality as
$$X^{(\mu \nu)} Y_{\mu \nu} = X^{(\mu \nu)} Y_{(\mu \nu)} \text{ .}$$
A: Since we have $X^{(\mu\nu)}=X^{(\nu\mu)}$ then
\begin{eqnarray*}
X^{(\mu\nu)}Y_{(\mu\nu)}&=&
\frac{1}{2}[X^{(\mu\nu)}Y_{\mu\nu}+X^{(\mu\nu)}Y_{\nu\mu}],\\
&=&\frac{1}{2}[X^{(\mu\nu)}Y_{\mu\nu}+X^{(\nu\mu)}Y_{\nu\mu}],\\
&=&\frac{1}{2}[X^{(\mu\nu)}Y_{\mu\nu}+X^{(\mu\nu)}Y_{\mu\nu}],\ \mbox{(because we label $\mu\leftrightarrow\nu$ the 2nd term})\\
&=&X^{(\mu\nu)}Y_{\mu\nu}.\\
\end{eqnarray*}
